Derivatives - AP Calculus AB (2024)

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AP Calculus AB Help » Derivatives

Example Question #1 : Derivatives

Differentiate,

Derivatives - AP Calculus AB (1)

Possible Answers:

Derivatives - AP Calculus AB (2)

Derivatives - AP Calculus AB (3)

Derivatives - AP Calculus AB (4)

Derivatives - AP Calculus AB (5)

Derivatives - AP Calculus AB (6)

Correct answer:

Derivatives - AP Calculus AB (7)

Explanation:

Differentiate,

Derivatives - AP Calculus AB (8)

Strategy

This one at first glance appears difficult even if we recognize that the chain rule is needed; we have a function within a function within a function within a function. To avoid making mistakes, it's best to start by defining variables to make the calculation easier to follow.

Let's start with the outermost function, we will writeDerivatives - AP Calculus AB (9)as a function ofDerivatives - AP Calculus AB (10)by setting,

Derivatives - AP Calculus AB (11)

______________________________________________________

Derivatives - AP Calculus AB (12)

Derivatives - AP Calculus AB (13)

_______________________________________________________

Similarly, define Derivatives - AP Calculus AB (14)to writeDerivatives - AP Calculus AB (15)as a function ofDerivatives - AP Calculus AB (16)

Derivatives - AP Calculus AB (17)

Derivatives - AP Calculus AB (18)

_______________________________________________________

WriteDerivatives - AP Calculus AB (19)as a function ofDerivatives - AP Calculus AB (20)

Derivatives - AP Calculus AB (21)

Derivatives - AP Calculus AB (22)

_______________________________________________________

Finally, define the inner-most function,Derivatives - AP Calculus AB (23),as the function ofDerivatives - AP Calculus AB (24)

Derivatives - AP Calculus AB (25)

________________________________________________________

Derivatives - AP Calculus AB (26)

SinceDerivatives - AP Calculus AB (27)we will just substitute that in and move to the front.

Derivatives - AP Calculus AB (28)

That was easy enough, now just write everything in terms ofDerivatives - AP Calculus AB (29)by going back to the definitions ofDerivatives - AP Calculus AB (30)Derivatives - AP Calculus AB (31)andDerivatives - AP Calculus AB (32).

Derivatives - AP Calculus AB (33)

Derivatives - AP Calculus AB (34)

Derivatives - AP Calculus AB (35)

Derivatives - AP Calculus AB (36)

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Example Question #2 : Derivatives

Find the tangent line. Given the point (1,2)

Derivatives - AP Calculus AB (37)

Possible Answers:

Derivatives - AP Calculus AB (38)

Derivatives - AP Calculus AB (39)

Derivatives - AP Calculus AB (40)

Derivatives - AP Calculus AB (41)

Derivatives - AP Calculus AB (42)

Correct answer:

Derivatives - AP Calculus AB (43)

Explanation:

To find the tangent line at the given point, we need to first take the derivative of the given function.

Power Rule:

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent.

Therefore,Derivatives - AP Calculus AB (44)becomesDerivatives - AP Calculus AB (45)

From there we plug in the "1" from the point to get our m value of the equationDerivatives - AP Calculus AB (46). When we plug in "1" to y' we get m=-1. Then from there, we will plug our point intoDerivatives - AP Calculus AB (47)now that we have found m to find our b value. So,

Derivatives - AP Calculus AB (48)

Derivatives - AP Calculus AB (49)

Therefore, the tangent line is equal to

Derivatives - AP Calculus AB (50)

Example Question #3 : Derivatives

Find the line tangent at the point (0,1)

Derivatives - AP Calculus AB (51)

Possible Answers:

Derivatives - AP Calculus AB (52)

Derivatives - AP Calculus AB (53)

Derivatives - AP Calculus AB (54)

Derivatives - AP Calculus AB (55)

Derivatives - AP Calculus AB (56)

Correct answer:

Derivatives - AP Calculus AB (57)

Explanation:

To find the tangent line at the given point, we need to first take the derivative of the given function. The rule for functions with "e" in it says that the derivative ofDerivatives - AP Calculus AB (58)However with this function there is also a 3 in the exponent so we will also use chain rule. Chain Rules states that we work from the outside to the inside. Meaning we will take the derivative of the outside of the equation and multiply it by the derivative of the inside of the equation.

To put this into equation it will look likeDerivatives - AP Calculus AB (59)

From there we plug in the "0" from the point to get our m value of the equation. When we plug in "0" to y' we get m=3. Then from there, we will plug our point intoDerivatives - AP Calculus AB (60)now that we have found m to find our m value. So,

Derivatives - AP Calculus AB (61)

Derivatives - AP Calculus AB (62)then plug this all back into the equation once more and we are left with

Derivatives - AP Calculus AB (63)

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Example Question #1 : Derivatives

Find the tangent line given the point (2,4) and the equation

Derivatives - AP Calculus AB (64)

Possible Answers:

Derivatives - AP Calculus AB (65)

Derivatives - AP Calculus AB (66)

Derivatives - AP Calculus AB (67)

Derivatives - AP Calculus AB (68)

Derivatives - AP Calculus AB (69)

Correct answer:

Derivatives - AP Calculus AB (70)

Explanation:

To find the tangent line at the given point, we need to first take the derivative of the given function using Power Rule

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent.Derivatives - AP Calculus AB (71)

From there we plug in the "2" from the point to get our m value of the equation. When we plug in "2" to y' we get m=8. Then from there, we will plug our point intonow that we have found m to find our bvalue. So,

Derivatives - AP Calculus AB (72)

Derivatives - AP Calculus AB (73)

Plug this back intoDerivatives - AP Calculus AB (74)

Derivatives - AP Calculus AB (75)

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Example Question #5 : Derivatives

Find the equation of the line tangent to the curve at the point whereDerivatives - AP Calculus AB (76)

Derivatives - AP Calculus AB (77)

Possible Answers:

Derivatives - AP Calculus AB (78)

Derivatives - AP Calculus AB (79)

Derivatives - AP Calculus AB (80)

Derivatives - AP Calculus AB (81)

Derivatives - AP Calculus AB (82)

Correct answer:

Derivatives - AP Calculus AB (83)

Explanation:

Find the equation of the line tangent to the curve Derivatives - AP Calculus AB (84)at the given point

Derivatives - AP Calculus AB (85) Derivatives - AP Calculus AB (86)

The slope of the line tangent at the given pointwill be equal to the derivative of Derivatives - AP Calculus AB (87)at that point. Compute the derivative and find the slope for our line:

Derivatives - AP Calculus AB (88)

Derivatives - AP Calculus AB (89)

Evaluate the secant term:

Derivatives - AP Calculus AB (90)

Therefore slope of the tangent line is simply:

Derivatives - AP Calculus AB (91)

So now we know the slope of the tangent line and can write the equation then solve for Derivatives - AP Calculus AB (92)

Derivatives - AP Calculus AB (93)

In order to solve for Derivatives - AP Calculus AB (94)we need one point on the line. Use the point where the tangent line meets the curve. Use the original function to find the "y" coordinate at this point:

Derivatives - AP Calculus AB (95)

We now have our point:

Derivatives - AP Calculus AB (96)

Use the point to findDerivatives - AP Calculus AB (97)

Derivatives - AP Calculus AB (98)

Derivatives - AP Calculus AB (99)

Derivatives - AP Calculus AB (100)

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Example Question #6 : Derivatives

Find the slope of the line tangent to the curve of d(g) when g=6.

Derivatives - AP Calculus AB (101)

Possible Answers:

Derivatives - AP Calculus AB (102)

Derivatives - AP Calculus AB (103)

Derivatives - AP Calculus AB (104)

Derivatives - AP Calculus AB (105)

Correct answer:

Derivatives - AP Calculus AB (106)

Explanation:

Find the slope of the line tangent to the curve of d(g) when g=6.

Derivatives - AP Calculus AB (107)

All we need here is the power rule. This states that to find the derivative of a polynomial, simply subtract 1 from each exponent and then multiply each term by their originalexponent.

Constant terms will drop out when we do this, and linear terms will become constants.

Derivatives - AP Calculus AB (108)

Derivatives - AP Calculus AB (109)

Derivatives - AP Calculus AB (110)

From here substitute in g=6.

Derivatives - AP Calculus AB (111)

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Example Question #1 : Derivative At A Point

Give the equation of the line tangent to the graph of the equation

Derivatives - AP Calculus AB (112)

at the pointDerivatives - AP Calculus AB (113).

Possible Answers:

Derivatives - AP Calculus AB (114)

Derivatives - AP Calculus AB (115)

Derivatives - AP Calculus AB (116)

Derivatives - AP Calculus AB (117)

Derivatives - AP Calculus AB (118)

Correct answer:

Derivatives - AP Calculus AB (119)

Explanation:

The tangent line to the graph ofDerivatives - AP Calculus AB (120)at pointDerivatives - AP Calculus AB (121)is the line with slopeDerivatives - AP Calculus AB (122)that passes through that point. Findthe derivativeDerivatives - AP Calculus AB (123):

Derivatives - AP Calculus AB (124)

Derivatives - AP Calculus AB (125)

Derivatives - AP Calculus AB (126)

Apply the sum rule:

Derivatives - AP Calculus AB (127)

Derivatives - AP Calculus AB (128)

Derivatives - AP Calculus AB (129)

Derivatives - AP Calculus AB (130)

Derivatives - AP Calculus AB (131)

The tangent line is therefore the line with slope 5 throughDerivatives - AP Calculus AB (132).Apply the point-slope formula:

Derivatives - AP Calculus AB (133)

Derivatives - AP Calculus AB (134)

Derivatives - AP Calculus AB (135)

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Example Question #8 : Derivatives

Give the equation of the line tangent to the graph of the equation

Derivatives - AP Calculus AB (136)

at the pointDerivatives - AP Calculus AB (137).

Possible Answers:

None of the other choices gives the correct response.

Derivatives - AP Calculus AB (138)

Derivatives - AP Calculus AB (139)

Derivatives - AP Calculus AB (140)

Derivatives - AP Calculus AB (141)

Correct answer:

None of the other choices gives the correct response.

Explanation:

The tangent line to the graph ofDerivatives - AP Calculus AB (142)at pointDerivatives - AP Calculus AB (143)is the line with slopeDerivatives - AP Calculus AB (144)that passes through that point. Findthe derivativeDerivatives - AP Calculus AB (145):

Derivatives - AP Calculus AB (146)

Derivatives - AP Calculus AB (147)

Apply the constant multiple and sum rules:

Derivatives - AP Calculus AB (148)

Derivatives - AP Calculus AB (149)

Derivatives - AP Calculus AB (150)

SetDerivatives - AP Calculus AB (151)andDerivatives - AP Calculus AB (152)and apply the chain rule.

Derivatives - AP Calculus AB (153)

Derivatives - AP Calculus AB (154)

Derivatives - AP Calculus AB (155)

Substituting back:

Derivatives - AP Calculus AB (156)

Derivatives - AP Calculus AB (157)

Derivatives - AP Calculus AB (158)

EvaluateDerivatives - AP Calculus AB (159)using substitution:

Derivatives - AP Calculus AB (160)

Derivatives - AP Calculus AB (161)

Derivatives - AP Calculus AB (162)

Derivatives - AP Calculus AB (163)

The tangent line is therefore the line with slope Derivatives - AP Calculus AB (164)throughDerivatives - AP Calculus AB (165).Derivatives - AP Calculus AB (166)is aDerivatives - AP Calculus AB (167)-intercept, so apply the slope-intercept formula to get the equation

Derivatives - AP Calculus AB (168).

This is not among the choices given.

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Example Question #1 : Ap Calculus Ab

Find the equation of the line parallel to the functionDerivatives - AP Calculus AB (169)atDerivatives - AP Calculus AB (170), and passes through the pointDerivatives - AP Calculus AB (171)

Possible Answers:

Derivatives - AP Calculus AB (172)

Derivatives - AP Calculus AB (173)

Derivatives - AP Calculus AB (174)

Derivatives - AP Calculus AB (175)

Correct answer:

Derivatives - AP Calculus AB (176)

Explanation:

We first start by finding the slope of the line in question, which we do by taking the derivative ofDerivatives - AP Calculus AB (177)and evaluate atDerivatives - AP Calculus AB (178).

Derivatives - AP Calculus AB (179),Derivatives - AP Calculus AB (180)

We then use point slope form to get the equation of the line at the pointDerivatives - AP Calculus AB (181)

Derivatives - AP Calculus AB (182)

Derivatives - AP Calculus AB (183)

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Example Question #10 : Derivatives

Find the equation of the line tangent toDerivatives - AP Calculus AB (184)at the point Derivatives - AP Calculus AB (185).

Possible Answers:

Derivatives - AP Calculus AB (186)

Derivatives - AP Calculus AB (187)

Derivatives - AP Calculus AB (188)

Derivatives - AP Calculus AB (189)

Derivatives - AP Calculus AB (190)

Correct answer:

Derivatives - AP Calculus AB (191)

Explanation:

The first step is to find the derivative of the function given, which isDerivatives - AP Calculus AB (192). Next, find the slope at (1,4) by plugging in x=1 and solving for Derivatives - AP Calculus AB (193), which is the slope. You should getDerivatives - AP Calculus AB (194). This means the slope of the new line is also -1 because at the point where a slope and a line are tangent they have the same slope. Use the equationDerivatives - AP Calculus AB (195) to express your line. Y and x are variables and mis the slope, so the only thing you need to find is b. Plug in the point and slope into Derivatives - AP Calculus AB (196)to get Derivatives - AP Calculus AB (197). Now you can express the general equation of the line as Derivatives - AP Calculus AB (198).

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Derivatives - AP Calculus AB (2024)

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